TELECOM CUSTOMER CHURN PREDICTION

Introduction

Customer Churn refers to the phenomenon where customers stop using a company's products or services.
In the telecom industry, churn is a major concern because retaining existing customers is significantly more cost-effective than acquiring new ones.

Telecom companies generate massive amounts of data, including customer demographics, account information, service usage patterns, and contract details.
By analyzing this data, we can build predictive models to identify customers who are likely to churn, allowing companies to take proactive measures to retain them.

In this project, we aim to:

• Understand the key factors influencing customer churn
• Explore the dataset using Exploratory Data Analysis (EDA)
• Apply Machine Learning algorithms to predict churn
• Evaluate model performance and provide actionable insights for customer retention

The main objectives of this project are:

1. Understand Customer Behavior
   Analyze customer demographics, account details, and service usage to identify patterns that influence churn.

2. Identify Key Churn Factors
   Determine which factors most strongly contribute to customer churn, such as contract type, payment method, or usage frequency.

3. Predict Customer Churn
   Build and evaluate machine learning models (e.g., Random Forest, Logistic Regression, SVM) to predict which customers are likely to leave.

4. Support Decision-Making
   Provide actionable insights that help telecom companies implement strategies to retain customers and reduce churn.

5. Evaluate Model Performance
   Measure accuracy, precision, recall, and other metrics to select the most effective predictive model.

Import Libraries

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import missingno as msno
import warnings
warnings.filterwarnings('ignore')
import plotly.express as px
import plotly.graph_objects as go
from plotly.subplots import make_subplots

from sklearn.preprocessing import StandardScaler, LabelEncoder
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestClassifier
from sklearn.tree import DecisionTreeClassifier
from sklearn.metrics import accuracy_score, confusion_matrix, classification_report

Load the dataset

df=pd.read_csv("chrun.csv")
print(df)

      customerID  gender  SeniorCitizen Partner Dependents  tenure  \
0     7590-VHVEG  Female              0     Yes         No       1   
1     5575-GNVDE    Male              0      No         No      34   
2     3668-QPYBK    Male              0      No         No       2   
3     7795-CFOCW    Male              0      No         No      45   
4     9237-HQITU  Female              0      No         No       2   
...          ...     ...            ...     ...        ...     ...   
7038  6840-RESVB    Male              0     Yes        Yes      24   
7039  2234-XADUH  Female              0     Yes        Yes      72   
7040  4801-JZAZL  Female              0     Yes        Yes      11   
7041  8361-LTMKD    Male              1     Yes         No       4   
7042  3186-AJIEK    Male              0      No         No      66   

     PhoneService     MultipleLines InternetService OnlineSecurity  ...  \
0              No  No phone service             DSL             No  ...   
1             Yes                No             DSL            Yes  ...   
2             Yes                No             DSL            Yes  ...   
3              No  No phone service             DSL            Yes  ...   
4             Yes                No     Fiber optic             No  ...   
...           ...               ...             ...            ...  ...   
7038          Yes               Yes             DSL            Yes  ...   
7039          Yes               Yes     Fiber optic             No  ...   
7040           No  No phone service             DSL            Yes  ...   
7041          Yes               Yes     Fiber optic             No  ...   
7042          Yes                No     Fiber optic            Yes  ...   

     DeviceProtection TechSupport StreamingTV StreamingMovies        Contract  \
0                  No          No          No              No  Month-to-month   
1                 Yes          No          No              No        One year   
2                  No          No          No              No  Month-to-month   
3                 Yes         Yes          No              No        One year   
4                  No          No          No              No  Month-to-month   
...               ...         ...         ...             ...             ...   
7038              Yes         Yes         Yes             Yes        One year   
7039              Yes          No         Yes             Yes        One year   
7040               No          No          No              No  Month-to-month   
7041               No          No          No              No  Month-to-month   
7042              Yes         Yes         Yes             Yes        Two year   

     PaperlessBilling              PaymentMethod MonthlyCharges  TotalCharges  \
0                 Yes           Electronic check          29.85         29.85   
1                  No               Mailed check          56.95        1889.5   
2                 Yes               Mailed check          53.85        108.15   
3                  No  Bank transfer (automatic)          42.30       1840.75   
4                 Yes           Electronic check          70.70        151.65   
...               ...                        ...            ...           ...   
7038              Yes               Mailed check          84.80        1990.5   
7039              Yes    Credit card (automatic)         103.20        7362.9   
7040              Yes           Electronic check          29.60        346.45   
7041              Yes               Mailed check          74.40         306.6   
7042              Yes  Bank transfer (automatic)         105.65        6844.5   

     Churn  
0       No  
1       No  
2      Yes  
3       No  
4      Yes  
...    ...  
7038    No  
7039    No  
7040    No  
7041   Yes  
7042    No  

[7043 rows x 21 columns]

df.head(10)

	customerID	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	...	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	7590-VHVEG	Female	0	Yes	No	1	No	No phone service	DSL	No	...	No	No	No	No	Month-to-month	Yes	Electronic check	29.85	29.85	No
1	5575-GNVDE	Male	0	No	No	34	Yes	No	DSL	Yes	...	Yes	No	No	No	One year	No	Mailed check	56.95	1889.5	No
2	3668-QPYBK	Male	0	No	No	2	Yes	No	DSL	Yes	...	No	No	No	No	Month-to-month	Yes	Mailed check	53.85	108.15	Yes
3	7795-CFOCW	Male	0	No	No	45	No	No phone service	DSL	Yes	...	Yes	Yes	No	No	One year	No	Bank transfer (automatic)	42.30	1840.75	No
4	9237-HQITU	Female	0	No	No	2	Yes	No	Fiber optic	No	...	No	No	No	No	Month-to-month	Yes	Electronic check	70.70	151.65	Yes
5	9305-CDSKC	Female	0	No	No	8	Yes	Yes	Fiber optic	No	...	Yes	No	Yes	Yes	Month-to-month	Yes	Electronic check	99.65	820.5	Yes
6	1452-KIOVK	Male	0	No	Yes	22	Yes	Yes	Fiber optic	No	...	No	No	Yes	No	Month-to-month	Yes	Credit card (automatic)	89.10	1949.4	No
7	6713-OKOMC	Female	0	No	No	10	No	No phone service	DSL	Yes	...	No	No	No	No	Month-to-month	No	Mailed check	29.75	301.9	No
8	7892-POOKP	Female	0	Yes	No	28	Yes	Yes	Fiber optic	No	...	Yes	Yes	Yes	Yes	Month-to-month	Yes	Electronic check	104.80	3046.05	Yes
9	6388-TABGU	Male	0	No	Yes	62	Yes	No	DSL	Yes	...	No	No	No	No	One year	No	Bank transfer (automatic)	56.15	3487.95	No

10 rows × 21 columns

The data set includes information about:

Customers who left within the last month – the column is called Churn

Services that each customer has signed up for – phone, multiple lines, internet, online security, online backup, device protection, tech support, and streaming TV and movies

Customer account information - how long they’ve been a customer, contract, payment method, paperless billing, monthly charges, and total charges

Demographic info about customers – gender, age range, and if they have partners and dependents

df.shape

(7043, 21)

df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 7043 entries, 0 to 7042
Data columns (total 21 columns):
 #   Column            Non-Null Count  Dtype  
---  ------            --------------  -----  
 0   customerID        7043 non-null   object 
 1   gender            7043 non-null   object 
 2   SeniorCitizen     7043 non-null   int64  
 3   Partner           7043 non-null   object 
 4   Dependents        7043 non-null   object 
 5   tenure            7043 non-null   int64  
 6   PhoneService      7043 non-null   object 
 7   MultipleLines     7043 non-null   object 
 8   InternetService   7043 non-null   object 
 9   OnlineSecurity    7043 non-null   object 
 10  OnlineBackup      7043 non-null   object 
 11  DeviceProtection  7043 non-null   object 
 12  TechSupport       7043 non-null   object 
 13  StreamingTV       7043 non-null   object 
 14  StreamingMovies   7043 non-null   object 
 15  Contract          7043 non-null   object 
 16  PaperlessBilling  7043 non-null   object 
 17  PaymentMethod     7043 non-null   object 
 18  MonthlyCharges    7043 non-null   float64
 19  TotalCharges      7043 non-null   object 
 20  Churn             7043 non-null   object 
dtypes: float64(1), int64(2), object(18)
memory usage: 1.1+ MB

df.columns.values

array(['customerID', 'gender', 'SeniorCitizen', 'Partner', 'Dependents',
       'tenure', 'PhoneService', 'MultipleLines', 'InternetService',
       'OnlineSecurity', 'OnlineBackup', 'DeviceProtection',
       'TechSupport', 'StreamingTV', 'StreamingMovies', 'Contract',
       'PaperlessBilling', 'PaymentMethod', 'MonthlyCharges',
       'TotalCharges', 'Churn'], dtype=object)

df.dtypes

customerID           object
gender               object
SeniorCitizen         int64
Partner              object
Dependents           object
tenure                int64
PhoneService         object
MultipleLines        object
InternetService      object
OnlineSecurity       object
OnlineBackup         object
DeviceProtection     object
TechSupport          object
StreamingTV          object
StreamingMovies      object
Contract             object
PaperlessBilling     object
PaymentMethod        object
MonthlyCharges      float64
TotalCharges         object
Churn                object
dtype: object

Visualize the missing value

msno.matrix(df)

<Axes: >

Data Manipulation

df=df.drop(['customerID'],axis=1)

df

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	Female	0	Yes	No	1	No	No phone service	DSL	No	Yes	No	No	No	No	Month-to-month	Yes	Electronic check	29.85	29.85	No
1	Male	0	No	No	34	Yes	No	DSL	Yes	No	Yes	No	No	No	One year	No	Mailed check	56.95	1889.5	No
2	Male	0	No	No	2	Yes	No	DSL	Yes	Yes	No	No	No	No	Month-to-month	Yes	Mailed check	53.85	108.15	Yes
3	Male	0	No	No	45	No	No phone service	DSL	Yes	No	Yes	Yes	No	No	One year	No	Bank transfer (automatic)	42.30	1840.75	No
4	Female	0	No	No	2	Yes	No	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Electronic check	70.70	151.65	Yes
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
7038	Male	0	Yes	Yes	24	Yes	Yes	DSL	Yes	No	Yes	Yes	Yes	Yes	One year	Yes	Mailed check	84.80	1990.5	No
7039	Female	0	Yes	Yes	72	Yes	Yes	Fiber optic	No	Yes	Yes	No	Yes	Yes	One year	Yes	Credit card (automatic)	103.20	7362.9	No
7040	Female	0	Yes	Yes	11	No	No phone service	DSL	Yes	No	No	No	No	No	Month-to-month	Yes	Electronic check	29.60	346.45	No
7041	Male	1	Yes	No	4	Yes	Yes	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Mailed check	74.40	306.6	Yes
7042	Male	0	No	No	66	Yes	No	Fiber optic	Yes	No	Yes	Yes	Yes	Yes	Two year	Yes	Bank transfer (automatic)	105.65	6844.5	No

7043 rows × 20 columns

On deep analysis, we can find some indirect missingness in our data (which can be in form of blankspaces). Let's see that!

df['TotalCharges']=pd.to_numeric(df.TotalCharges,errors='coerce')

df.dtypes

gender               object
SeniorCitizen         int64
Partner              object
Dependents           object
tenure                int64
PhoneService         object
MultipleLines        object
InternetService      object
OnlineSecurity       object
OnlineBackup         object
DeviceProtection     object
TechSupport          object
StreamingTV          object
StreamingMovies      object
Contract             object
PaperlessBilling     object
PaymentMethod        object
MonthlyCharges      float64
TotalCharges        float64
Churn                object
dtype: object

df.isnull().sum()

gender               0
SeniorCitizen        0
Partner              0
Dependents           0
tenure               0
PhoneService         0
MultipleLines        0
InternetService      0
OnlineSecurity       0
OnlineBackup         0
DeviceProtection     0
TechSupport          0
StreamingTV          0
StreamingMovies      0
Contract             0
PaperlessBilling     0
PaymentMethod        0
MonthlyCharges       0
TotalCharges        11
Churn                0
dtype: int64

msno.matrix(df)

<Axes: >

df[np.isnan(df['TotalCharges'])]

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
488	Female	0	Yes	Yes	0	No	No phone service	DSL	Yes	No	Yes	Yes	Yes	No	Two year	Yes	Bank transfer (automatic)	52.55	NaN	No
753	Male	0	No	Yes	0	Yes	No	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	Two year	No	Mailed check	20.25	NaN	No
936	Female	0	Yes	Yes	0	Yes	No	DSL	Yes	Yes	Yes	No	Yes	Yes	Two year	No	Mailed check	80.85	NaN	No
1082	Male	0	Yes	Yes	0	Yes	Yes	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	Two year	No	Mailed check	25.75	NaN	No
1340	Female	0	Yes	Yes	0	No	No phone service	DSL	Yes	Yes	Yes	Yes	Yes	No	Two year	No	Credit card (automatic)	56.05	NaN	No
3331	Male	0	Yes	Yes	0	Yes	No	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	Two year	No	Mailed check	19.85	NaN	No
3826	Male	0	Yes	Yes	0	Yes	Yes	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	Two year	No	Mailed check	25.35	NaN	No
4380	Female	0	Yes	Yes	0	Yes	No	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	Two year	No	Mailed check	20.00	NaN	No
5218	Male	0	Yes	Yes	0	Yes	No	No	No internet service	No internet service	No internet service	No internet service	No internet service	No internet service	One year	Yes	Mailed check	19.70	NaN	No
6670	Female	0	Yes	Yes	0	Yes	Yes	DSL	No	Yes	Yes	Yes	Yes	No	Two year	No	Mailed check	73.35	NaN	No
6754	Male	0	No	Yes	0	Yes	Yes	DSL	Yes	Yes	No	Yes	No	No	Two year	Yes	Bank transfer (automatic)	61.90	NaN	No

It can also be noted that the Tenure column is 0 for these entries even though the MonthlyCharges column is not empty. Let's see if there are any other 0 values in the tenure column.

df[df['tenure']==0].index

Index([488, 753, 936, 1082, 1340, 3331, 3826, 4380, 5218, 6670, 6754], dtype='int64')

There are no additional missing values in the Tenure column. Let's delete the rows with missing values in Tenure columns since there are only 11 rows and deleting them will not affect the data.

df.drop(labels=df[df['tenure']==0].index,axis=0,inplace=True)

df

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	Female	0	Yes	No	1	No	No phone service	DSL	No	Yes	No	No	No	No	Month-to-month	Yes	Electronic check	29.85	29.85	No
1	Male	0	No	No	34	Yes	No	DSL	Yes	No	Yes	No	No	No	One year	No	Mailed check	56.95	1889.50	No
2	Male	0	No	No	2	Yes	No	DSL	Yes	Yes	No	No	No	No	Month-to-month	Yes	Mailed check	53.85	108.15	Yes
3	Male	0	No	No	45	No	No phone service	DSL	Yes	No	Yes	Yes	No	No	One year	No	Bank transfer (automatic)	42.30	1840.75	No
4	Female	0	No	No	2	Yes	No	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Electronic check	70.70	151.65	Yes
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
7038	Male	0	Yes	Yes	24	Yes	Yes	DSL	Yes	No	Yes	Yes	Yes	Yes	One year	Yes	Mailed check	84.80	1990.50	No
7039	Female	0	Yes	Yes	72	Yes	Yes	Fiber optic	No	Yes	Yes	No	Yes	Yes	One year	Yes	Credit card (automatic)	103.20	7362.90	No
7040	Female	0	Yes	Yes	11	No	No phone service	DSL	Yes	No	No	No	No	No	Month-to-month	Yes	Electronic check	29.60	346.45	No
7041	Male	1	Yes	No	4	Yes	Yes	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Mailed check	74.40	306.60	Yes
7042	Male	0	No	No	66	Yes	No	Fiber optic	Yes	No	Yes	Yes	Yes	Yes	Two year	Yes	Bank transfer (automatic)	105.65	6844.50	No

7032 rows × 20 columns

df.skew(numeric_only=True)

SeniorCitizen     1.831103
tenure            0.237731
MonthlyCharges   -0.222103
TotalCharges      0.961642
dtype: float64

To solve the problem of missing values in TotalCharges column, I decided to fill it with the mean of TotalCharges values.

df.fillna(df['TotalCharges'].mean())

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	Female	0	Yes	No	1	No	No phone service	DSL	No	Yes	No	No	No	No	Month-to-month	Yes	Electronic check	29.85	29.85	No
1	Male	0	No	No	34	Yes	No	DSL	Yes	No	Yes	No	No	No	One year	No	Mailed check	56.95	1889.50	No
2	Male	0	No	No	2	Yes	No	DSL	Yes	Yes	No	No	No	No	Month-to-month	Yes	Mailed check	53.85	108.15	Yes
3	Male	0	No	No	45	No	No phone service	DSL	Yes	No	Yes	Yes	No	No	One year	No	Bank transfer (automatic)	42.30	1840.75	No
4	Female	0	No	No	2	Yes	No	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Electronic check	70.70	151.65	Yes
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
7038	Male	0	Yes	Yes	24	Yes	Yes	DSL	Yes	No	Yes	Yes	Yes	Yes	One year	Yes	Mailed check	84.80	1990.50	No
7039	Female	0	Yes	Yes	72	Yes	Yes	Fiber optic	No	Yes	Yes	No	Yes	Yes	One year	Yes	Credit card (automatic)	103.20	7362.90	No
7040	Female	0	Yes	Yes	11	No	No phone service	DSL	Yes	No	No	No	No	No	Month-to-month	Yes	Electronic check	29.60	346.45	No
7041	Male	1	Yes	No	4	Yes	Yes	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Mailed check	74.40	306.60	Yes
7042	Male	0	No	No	66	Yes	No	Fiber optic	Yes	No	Yes	Yes	Yes	Yes	Two year	Yes	Bank transfer (automatic)	105.65	6844.50	No

7032 rows × 20 columns

df.isnull().sum()

gender              0
SeniorCitizen       0
Partner             0
Dependents          0
tenure              0
PhoneService        0
MultipleLines       0
InternetService     0
OnlineSecurity      0
OnlineBackup        0
DeviceProtection    0
TechSupport         0
StreamingTV         0
StreamingMovies     0
Contract            0
PaperlessBilling    0
PaymentMethod       0
MonthlyCharges      0
TotalCharges        0
Churn               0
dtype: int64

df['SeniorCitizen']=df['SeniorCitizen'].map({0:'No',1:'Yes'})

df

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	Female	No	Yes	No	1	No	No phone service	DSL	No	Yes	No	No	No	No	Month-to-month	Yes	Electronic check	29.85	29.85	No
1	Male	No	No	No	34	Yes	No	DSL	Yes	No	Yes	No	No	No	One year	No	Mailed check	56.95	1889.50	No
2	Male	No	No	No	2	Yes	No	DSL	Yes	Yes	No	No	No	No	Month-to-month	Yes	Mailed check	53.85	108.15	Yes
3	Male	No	No	No	45	No	No phone service	DSL	Yes	No	Yes	Yes	No	No	One year	No	Bank transfer (automatic)	42.30	1840.75	No
4	Female	No	No	No	2	Yes	No	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Electronic check	70.70	151.65	Yes
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
7038	Male	No	Yes	Yes	24	Yes	Yes	DSL	Yes	No	Yes	Yes	Yes	Yes	One year	Yes	Mailed check	84.80	1990.50	No
7039	Female	No	Yes	Yes	72	Yes	Yes	Fiber optic	No	Yes	Yes	No	Yes	Yes	One year	Yes	Credit card (automatic)	103.20	7362.90	No
7040	Female	No	Yes	Yes	11	No	No phone service	DSL	Yes	No	No	No	No	No	Month-to-month	Yes	Electronic check	29.60	346.45	No
7041	Male	Yes	Yes	No	4	Yes	Yes	Fiber optic	No	No	No	No	No	No	Month-to-month	Yes	Mailed check	74.40	306.60	Yes
7042	Male	No	No	No	66	Yes	No	Fiber optic	Yes	No	Yes	Yes	Yes	Yes	Two year	Yes	Bank transfer (automatic)	105.65	6844.50	No

7032 rows × 20 columns

df.nunique()

gender                 2
SeniorCitizen          2
Partner                2
Dependents             2
tenure                72
PhoneService           2
MultipleLines          3
InternetService        3
OnlineSecurity         3
OnlineBackup           3
DeviceProtection       3
TechSupport            3
StreamingTV            3
StreamingMovies        3
Contract               3
PaperlessBilling       2
PaymentMethod          4
MonthlyCharges      1584
TotalCharges        6530
Churn                  2
dtype: int64

df['InternetService'].describe()

count            7032
unique              3
top       Fiber optic
freq             3096
Name: InternetService, dtype: object

df.dtypes

gender               object
SeniorCitizen        object
Partner              object
Dependents           object
tenure                int64
PhoneService         object
MultipleLines        object
InternetService      object
OnlineSecurity       object
OnlineBackup         object
DeviceProtection     object
TechSupport          object
StreamingTV          object
StreamingMovies      object
Contract             object
PaperlessBilling     object
PaymentMethod        object
MonthlyCharges      float64
TotalCharges        float64
Churn                object
dtype: object

num=['tenure','MonthlyCharges','TotalCharges']
df[num].describe()

	tenure	MonthlyCharges	TotalCharges
count	7032.000000	7032.000000	7032.000000
mean	32.421786	64.798208	2283.300441
std	24.545260	30.085974	2266.771362
min	1.000000	18.250000	18.800000
25%	9.000000	35.587500	401.450000
50%	29.000000	70.350000	1397.475000
75%	55.000000	89.862500	3794.737500
max	72.000000	118.750000	8684.800000

Data Visualization

g_labels=['Male','Female']
c_labels=['No','Yes']
fig=make_subplots(rows=1,cols=2,specs=[[{'type':'domain'},{'type':'domain'}]])
fig.add_trace(go.Pie(labels=g_labels, values=df['gender'].value_counts(), name="Gender"),1,1)
fig.add_trace(go.Pie(labels=c_labels, values=df['Churn'].value_counts(),name='Churn'),1,2)
fig.update_traces(hole=.4,hoverinfo="label+percent+name",textfont_size=16)
fig.update_layout(
    title_text="<b>Gender and Churn Distribution<b>",
    annotations=[dict(text='Gender',x=0.19,y=0.5,font_size=16,showarrow=False),
               dict(text="Churn",x=0.8,y=0.5,font_size=16,showarrow=False)])
fig.show()

26.6 % of customers switched to another firm. The pie charts show that the dataset has an almost equal distribution of male and female customers. Overall churn percentage is smaller compared to customers who stayed. This suggests that gender does not have a strong impact on churn likelihood in this dataset.

df["Churn"][df["Churn"]=='No'].groupby(by=df['gender']).count()

gender
Female    2544
Male      2619
Name: Churn, dtype: int64

df["Churn"][df["Churn"]=='Yes'].groupby(by=df['gender']).count()

gender
Female    939
Male      930
Name: Churn, dtype: int64

plt.figure(figsize=(6,6))
labels=['Churn: Yes',"Churn: No"]
values=[1869,5163]
colors=["plum","wheat"]
labels_gender=["F","M","F","M"]
values_gender=[939,930,2544,2619]
colors_gender=['skyblue','lightpink','skyblue','lightpink']
explode=[0.3,0.3]
explode_gender=[0.1,0.1,0.1,0.1]
textprops={"fontsize":13}
textprops1={"fontsize":10}

plt.pie(
    values,
    labels=labels,
    autopct='%1.1f%%',
    pctdistance=1.08,
    labeldistance=0.8,
    colors=colors,
    startangle=90,
    frame=True,
    explode=explode,
    radius=10,
    textprops=textprops,
    counterclock=True
)
plt.pie(
    values_gender,
    labels=labels_gender,
    autopct='%1.1f%%',
    pctdistance=0.55,
    labeldistance=0.82,
    colors=colors_gender,
    startangle=90,
    explode=explode_gender,
    radius=7,
    textprops=textprops1,
    counterclock=True
)

centre_circle=plt.Circle((0,0),5,color='black',fc='white',linewidth=0)
fig=plt.gcf()
fig.gca().add_artist(centre_circle)

plt.title("Churn distribution w.r.t Gender: Male(M) , Female(F)",fontsize=11,y=1.1)

plt.axis('equal')
plt.tight_layout()
plt.legend(fontsize=8)
plt.show()

Both males and females have similar churn proportions. This indicates that customer gender is not a major factor influencing whether they leave the service. Hence, churn behavior appears consistent across both genders.

df.dtypes

gender               object
SeniorCitizen        object
Partner              object
Dependents           object
tenure                int64
PhoneService         object
MultipleLines        object
InternetService      object
OnlineSecurity       object
OnlineBackup         object
DeviceProtection     object
TechSupport          object
StreamingTV          object
StreamingMovies      object
Contract             object
PaperlessBilling     object
PaymentMethod        object
MonthlyCharges      float64
TotalCharges        float64
Churn                object
dtype: object

fig=px.histogram(df, x='Churn', color='Contract', barmode='group', title="<b>Customer Contract Distribution<b>")
fig.update_layout(width=700, height=500, bargap=0.1)
fig.show()

Month-to-month contracts show the highest churn rate, unlike one- or two-year contracts. Customers with long-term commitments tend to remain loyal to the service provider. Offering incentives for longer contracts can help minimize churn.

labels=df['PaymentMethod'].unique()
values=df['PaymentMethod'].value_counts()
fig=go.Figure(data=[go.Pie(labels=labels,values=values,hole=0.3)])
fig.update_layout(width=700, title="<b>Payment Method Distribution<b>")
fig.show()

fig=px.histogram(df, x='Churn', color='PaymentMethod', title="<b>Customer Payment distribution w.r.t Churn<b>")
fig.update_layout(width=700, height=500, bargap=0.1)
fig.show()

Customers paying through electronic checks show the highest churn. Those using credit cards or bank transfers remain more consistent customers. This may reflect convenience, automation, and lower billing issues with digital payments.

df['InternetService'].unique()

array(['DSL', 'Fiber optic', 'No'], dtype=object)

df[df["gender"]=="Female"][["InternetService","Churn"]].value_counts()

InternetService  Churn
DSL              No       965
Fiber optic      No       889
No               No       690
Fiber optic      Yes      664
DSL              Yes      219
No               Yes       56
Name: count, dtype: int64

df[df["gender"]=="Male"][["InternetService","Churn"]].value_counts()

InternetService  Churn
DSL              No       992
Fiber optic      No       910
No               No       717
Fiber optic      Yes      633
DSL              Yes      240
No               Yes       57
Name: count, dtype: int64

fig=go.Figure()

fig.add_trace(go.Bar(
    x=[['Churn:No', 'Churn:No', 'Churn:Yes', 'Churn:Yes'],
      ['Female', 'Male', 'Female', 'Male']],
    y=[965,992,219,240],
    name='DSL'
))
fig.add_trace(go.Bar(
    x=[['Churn:No', 'Churn:No', 'Churn:Yes', 'Churn:Yes'],
      ['Female', 'Male', 'Female', 'Male']],
    y=[889,910,664,633],
    name='Fibre Optic'
))
fig.add_trace(go.Bar(
    x=[['Churn:No', 'Churn:No', 'Churn:Yes', 'Churn:Yes'],
      ['Female', 'Male', 'Female', 'Male']],
    y=[690,717,717,57]
))

fig.update_layout(width=900, height=400, title="<b>Churn Distribution w.r.t to Internet services and Gender<b>")
fig.show()

A lot of customers choose the Fiber optic service and it's also evident that the customers who use Fiber optic have high churn rate, this might suggest a dissatisfaction with this type of internet service. Customers having DSL service are majority in number and have less churn rate compared to Fibre optic service.

color={"Yes":"violet","No":"cyan"}
fig=px.histogram(
    df,
    x='Churn',
    color='Dependents',
    barmode='group',
    title="<b>Dependents Distribution<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

Customers with partners or dependents tend to stay longer and churn less frequently. This could be because family users rely more on stable communication services. Hence, single or independent customers are more at risk of leaving.

color={"Yes":"hotpink","No":"lightblue"}
fig=px.histogram(
    df,
    x="Churn",
    color="Partner",
    barmode="group",
    title="<b>Churn Distribution w.r.t Partner<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

color={"Yes":"palegreen","No":"khaki"}
fig=px.histogram(
    df,
    x="Churn",
    color="SeniorCitizen",
    barmode="group",
    title="<b>Churn Distribution w.r.t Senior Citizen<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

Senior citizens exhibit a noticeably higher churn rate than non-senior customers. This suggests that age or technology comfort level may influence customer satisfaction. Older customers may require more personalized support to improve retention.

color={"Yes":"deeppink","No":"darkviolet","No internet service":"lightgreen"}
fig=px.histogram(
    df,
    x="Churn",
    color="OnlineSecurity",
    barmode="group",
    title="<b>Churn Distribution w.r.t Online Security<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

Customers lacking online security services churn more often than those who have them. It shows that value-added services like security increase satisfaction and retention. Encouraging such subscriptions could help in lowering churn rates.

color={"Yes":"deeppink","No":"darkviolet","No phone service":"lightgreen"}
fig=px.histogram(
    df,
    x="Churn",
    color="MultipleLines",
    barmode="group",
    title="<b>Churn Distribution w.r.t MultipleLines<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

Customers with multiple lines or bundled services tend to churn less. This indicates that bundled or multi-service offerings increase stickiness. Cross-selling more services can be an effective churn reduction strategy.

color={"Yes":"black","No":"red"}
fig=px.histogram(
    df,
    x="Churn",
    color="PaperlessBilling",
    barmode="group",
    title="<b>Churn Distribution w.r.t PaperlessBilling<b>",
    color_discrete_map=color
)
fig.update_layout(
    width=700,
    height=500,
    bargap=0.1
)
fig.show()

Paperless billing users show higher churn rates than those using mailed bills. Such customers may be more digitally active and more likely to switch services. Targeted offers or engagement strategies for digital users could help retention.

fig=px.scatter(
    df,
    x="tenure",
    y="MonthlyCharges",
    color="Churn",
    title="<b>Tenure vs Monthly Charges By churn"
)
fig.update_layout(
    width=800,
    height=600
)
fig.show()

fig=px.scatter(
    df,
    x="tenure",
    y="MonthlyCharges",
    size="TotalCharges",
    color="Churn",
    title="<b>Customer value By Tenure and Charges<b>"
)
fig.update_layout(
    width=800,
    height=700
)
fig.show()

fig=px.box(
    df,
    x="Contract",
    y="MonthlyCharges",
    color="Churn",
    title="<b>Monthly Charges By Contract type and Churn<b>"
)
fig.update_layout(
    height=600
)
fig.show()

fig=px.violin(
    df,
    x="InternetService",
    y="tenure",
    color="Churn",
    title="<b>Tenure Distribution by Internet Services and Churn<b>"
)
fig.update_layout(
    height=600,
    
)
fig.show()

corr=df.corr(numeric_only=True)
fig=px.imshow(
    corr,
    text_auto=True,
    color_continuous_scale="PiYG",
    title="Correlation Heatmap"
)
fig.update_layout(
    width=700,
    height=500
)
fig.show()

sns.pairplot(df,vars=["tenure","MonthlyCharges","TotalCharges"],hue="Churn")

<seaborn.axisgrid.PairGrid at 0x1d61ec22e40>

Encoding the categorical columns

le=LabelEncoder()
for col in df.select_dtypes("object").columns:
    df[col]=le.fit_transform(df[col])
df    

	gender	SeniorCitizen	Partner	Dependents	tenure	PhoneService	MultipleLines	InternetService	OnlineSecurity	OnlineBackup	DeviceProtection	TechSupport	StreamingTV	StreamingMovies	Contract	PaperlessBilling	PaymentMethod	MonthlyCharges	TotalCharges	Churn
0	0	0	1	0	1	0	1	0	0	2	0	0	0	0	0	1	2	29.85	29.85	0
1	1	0	0	0	34	1	0	0	2	0	2	0	0	0	1	0	3	56.95	1889.50	0
2	1	0	0	0	2	1	0	0	2	2	0	0	0	0	0	1	3	53.85	108.15	1
3	1	0	0	0	45	0	1	0	2	0	2	2	0	0	1	0	0	42.30	1840.75	0
4	0	0	0	0	2	1	0	1	0	0	0	0	0	0	0	1	2	70.70	151.65	1
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
7038	1	0	1	1	24	1	2	0	2	0	2	2	2	2	1	1	3	84.80	1990.50	0
7039	0	0	1	1	72	1	2	1	0	2	2	0	2	2	1	1	1	103.20	7362.90	0
7040	0	0	1	1	11	0	1	0	2	0	0	0	0	0	0	1	2	29.60	346.45	0
7041	1	1	1	0	4	1	2	1	0	0	0	0	0	0	0	1	3	74.40	306.60	1
7042	1	0	0	0	66	1	0	1	2	0	2	2	2	2	2	1	0	105.65	6844.50	0

7032 rows × 20 columns

x=df.drop('Churn',axis=1)
y=df['Churn']

x_train,x_test,y_train,y_test=train_test_split(x,y,test_size=0.2,random_state=42)

Model Training

dt=DecisionTreeClassifier(criterion="entropy",random_state=42)
dt.fit(x_train,y_train)
dt_preds=dt.predict(x_test)

rf=RandomForestClassifier(random_state=42)
rf.fit(x_train,y_train)
rf_preds = rf.predict(x_test)

print("Decision Tree Accuracy:", accuracy_score(y_test, dt_preds))
print("Random Forest Accuracy:", accuracy_score(y_test, rf_preds))

Decision Tree Accuracy: 0.7448471926083866
Random Forest Accuracy: 0.7924662402274343

Here we can see that Random Forest model has more accuracy than the Decision Tree model.

print("DT Confusion Matrix:\n", confusion_matrix(y_test, dt_preds))
print("RF Confusion Matrix:\n", confusion_matrix(y_test, rf_preds))

DT Confusion Matrix:
 [[841 192]
 [167 207]]
RF Confusion Matrix:
 [[932 101]
 [191 183]]

Confusion Matrix

dt_conf=confusion_matrix(y_test, dt_preds)
sns.heatmap(
    dt_conf,
    annot=True,
    fmt='d',
    cmap='Blues',
    xticklabels=['No Churn','Churn'],
    yticklabels=['No Churn','Churn'],
)
plt.xlabel('Predicted')
plt.ylabel('Actual')
plt.title('Confusion Matrix of Decision Tree')
plt.show()

rf_conf=confusion_matrix(y_test, rf_preds)
sns.heatmap(
    rf_conf,
    annot=True,
    fmt='d',
    cmap='magma',
    xticklabels=['No Churn','Churn'],
    yticklabels=['No Churn','Churn'],
)
plt.xlabel('Predicted')
plt.ylabel('Actual')
plt.title('Confusion Matrix of Random Forest')
plt.show()

Classsification Reports

dt_report=classification_report(y_test,dt_preds)
print(dt_report)

              precision    recall  f1-score   support

           0       0.83      0.81      0.82      1033
           1       0.52      0.55      0.54       374

    accuracy                           0.74      1407
   macro avg       0.68      0.68      0.68      1407
weighted avg       0.75      0.74      0.75      1407

rf_report=classification_report(y_test,rf_preds)
print(rf_report)

              precision    recall  f1-score   support

           0       0.83      0.90      0.86      1033
           1       0.64      0.49      0.56       374

    accuracy                           0.79      1407
   macro avg       0.74      0.70      0.71      1407
weighted avg       0.78      0.79      0.78      1407